#### Transcript Chapter 4: Divide-and

Chapter 4 Divide-and-Conquer Copyright © 2007 Pearson Addison-Wesley. All rights reserved. Divide-and-Conquer The most-well known algorithm design strategy: 1. Divide instance of problem into two or more smaller instances 2. Solve smaller instances recursively 3. Obtain solution to original (larger) instance by combining these solutions Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-1 Divide-and-Conquer Technique (cont.) a problem of size n (instance) subproblem 1 of size n/2 subproblem 2 of size n/2 a solution to subproblem 1 a solution to subproblem 2 a solution to the original problem Copyright © 2007 Pearson Addison-Wesley. All rights reserved. It general leads to a recursive algorithm! A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-2 Divide-and-Conquer Examples Sorting: mergesort and quicksort Binary tree traversals Binary search (?) Multiplication of large integers Matrix multiplication: Strassen’s algorithm Closest-pair and convex-hull algorithms Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-3 General Divide-and-Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n) (nd), d 0 Master Theorem: If a < bd, T(n) (nd) If a = bd, T(n) (nd log n) If a > bd, T(n) (nlog b a ) Note: The same results hold with O instead of . Examples: T(n) = 4T(n/2) + n T(n) ? T(n) = 4T(n/2) + n2 T(n) ? T(n) = 4T(n/2) + n3 T(n) ? Copyright © 2007 Pearson Addison-Wesley. All rights reserved. (n^2) (n^2log n) (n^3) A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-4 Mergesort Split array A[0..n-1] into about equal halves and make copies of each half in arrays B and C Sort arrays B and C recursively Merge sorted arrays B and C into array A as follows: • Repeat the following until no elements remain in one of the arrays: – compare the first elements in the remaining unprocessed portions of the arrays – copy the smaller of the two into A, while incrementing the index indicating the unprocessed portion of that array • Once all elements in one of the arrays are processed, copy the remaining unprocessed elements from the other array into A. Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-5 Pseudocode of Mergesort Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-6 Pseudocode of Merge Time complexity: Θ(p+q) = Θ(n) comparisons Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-7 Mergesort Example 8 3 2 9 7 1 5 4 8 3 2 9 8 3 8 7 1 5 4 2 9 3 2 3 8 71 9 2 9 7 5 4 1 5 1 7 2 3 8 9 4 4 5 1 4 5 7 The non-recursive version of Mergesort starts from merging single elements into sorted pairs. 1 2 3 4 5 7 8 9 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-8 Analysis of Mergesort All cases have same efficiency: Θ(n log n) T(n) = 2T(n/2) + Θ(n), T(1) = 0 Number of comparisons in the worst case is close to theoretical minimum for comparison-based sorting: log2 n! ≈ n log2 n - 1.44n Space requirement: Θ(n) (not in-place) Can be implemented without recursion (bottom-up) Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-9 Quicksort Select a pivot (partitioning element) – here, the first element Rearrange the list so that all the elements in the first s positions are smaller than or equal to the pivot and all the elements in the remaining n-s positions are larger than or equal to the pivot (see next slide for an algorithm) p A[i]p A[i]p Exchange the pivot with the last element in the first (i.e., ) subarray — the pivot is now in its final position Sort the two subarrays recursively Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-10 Partitioning Algorithm < or i > r or j = l Time complexity: Θ(r-l) comparisons Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-11 Quicksort Example 5 3 1 9 8 2 4 7 2 3 1 4 5 8 9 7 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-12 Analysis of Quicksort Best case: split in the middle — Θ(n log n) Worst case: sorted array! — Θ(n2) T(n) = T(n-1) + Θ(n) Average case: random arrays — Θ(n log n) Improvements: • better pivot selection: median of three partitioning • switch to insertion sort on small subfiles • elimination of recursion These combine to 20-25% improvement Considered the method of choice for internal sorting of large files (n ≥ 10000) Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-13 Binary Search Very efficient algorithm for searching in sorted array: K vs A[0] . . . A[m] . . . A[n-1] If K = A[m], stop (successful search); otherwise, continue searching by the same method in A[0..m-1] if K < A[m] and in A[m+1..n-1] if K > A[m] l 0; r n-1 while l r do m (l+r)/2 if K = A[m] return m else if K < A[m] r m-1 else l m+1 return -1 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-14 Analysis of Binary Search Time efficiency • worst-case recurrence: Cw (n) = 1 + Cw( n/2 ), Cw (1) = 1 solution: Cw(n) = log2(n+1) This is VERY fast: e.g., Cw(106) = 20 Optimal for searching a sorted array Limitations: must be a sorted array (not linked list) Bad (degenerate) example of divide-and-conquer because only one of the sub-instances is solved Has a continuous counterpart called bisection method for solving equations in one unknown f(x) = 0 (see Sec. 12.4) Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-15 Binary Tree Algorithms Binary tree is a divide-and-conquer ready structure! Ex. 1: Classic traversals (preorder, inorder, postorder) Algorithm Inorder(T) if T Inorder(Tleft) print(root of T) Inorder(Tright) Efficiency: Θ(n). Why? Copyright © 2007 Pearson Addison-Wesley. All rights reserved. a b a c d e b c • • d e •••• Each node is visited/printed once. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-16 Binary Tree Algorithms (cont.) Ex. 2: Computing the height of a binary tree TL TR h(T) = max{h(TL), h(TR)} + 1 if T and h() = -1 Efficiency: Θ(n). Why? Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-17 Multiplication of Large Integers Consider the problem of multiplying two (large) n-digit integers represented by arrays of their digits such as: A = 12345678901357986429 B = 87654321284820912836 The grade-school algorithm: a1 a2 … an b1 b2 … bn (d10) d11d12 … d1n (d20) d21d22 … d2n ………………… (dn0) dn1dn2 … dnn Efficiency: Θ(n2) single-digit multiplications Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-18 First Divide-and-Conquer Algorithm A small example: A B where A = 2135 and B = 4014 A = (21·102 + 35), B = (40 ·102 + 14) So, A B = (21 ·102 + 35) (40 ·102 + 14) = 21 40 ·104 + (21 14 + 35 40) ·102 + 35 14 In general, if A = A1A2 and B = B1B2 (where A and B are n-digit, A1, A2, B1, B2 are n/2-digit numbers), A B = A1 B1·10n + (A1 B2 + A2 B1) ·10n/2 + A2 B2 Recurrence for the number of one-digit multiplications M(n): M(n) = 4M(n/2), M(1) = 1 Solution: M(n) = n2 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-19 Second Divide-and-Conquer Algorithm A B = A1 B1·10n + (A1 B2 + A2 B1) ·10n/2 + A2 B2 The idea is to decrease the number of multiplications from 4 to 3: (A1 + A2 ) (B1 + B2 ) = A1 B1 + (A1 B2 + A2 B1) + A2 B2, I.e., (A1 B2 + A2 B1) = (A1 + A2 ) (B1 + B2 ) - A1 B1 - A2 B2, which requires only 3 multiplications at the expense of (4-1) extra add/sub. Recurrence for the number of multiplications M(n): What if we count M(n) = 3M(n/2), M(1) = 1 both multiplications Solution: M(n) = 3log 2n = nlog 23 ≈ n1.585 and additions? Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-20 Example of Large-Integer Multiplication 2135 4014 = (21*10^2 + 35) * (40*10^2 + 14) = (21*40)*10^4 + c1*10^2 + 35*14 where c1 = (21+35)*(40+14) - 21*40 - 35*14, and 21*40 = (2*10 + 1) * (4*10 + 0) = (2*4)*10^2 + c2*10 + 1*0 where c2 = (2+1)*(4+0) - 2*4 - 1*0, etc. This process requires 9 digit multiplications as opposed to 16. Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-21 Conventional Matrix Multiplication Brute-force algorithm c00 c01 a00 a01 = * c10 c11 a10 a11 b00 b01 b10 b11 a00 * b00 + a01 * b10 a00 * b01 + a01 * b11 = a10 * b00 + a11 * b10 8 multiplications a10 * b01 + a11 * b11 Efficiency class in general: (n3) 4 additions Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-22 Strassen’s Matrix Multiplication Strassen’s algorithm for two 2x2 matrices (1969): c00 c01 a00 a01 b00 b01 = * c10 c11 a10 a11 b10 b11 m1 + m4 - m5 + m7 m3 + m5 = m2 + m4 m1 = (a00 + a11) * (b00 + b11) m2 = (a10 + a11) * b00 m3 = a00 * (b01 - b11) m4 = a11 * (b10 - b00) m5 = (a00 + a01) * b11 m6 = (a10 - a00) * (b00 + b01) m7 = (a01 - a11) * (b10 + b11) Copyright © 2007 Pearson Addison-Wesley. All rights reserved. m1 + m3 - m2 + m6 7 multiplications 18 additions A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-23 Strassen’s Matrix Multiplication Strassen observed [1969] that the product of two matrices can be computed in general as follows: C00 C01 A00 A01 = C10 C11 B00 B01 * A10 A11 B10 B11 M 1 + M 4 - M5 + M 7 M3 + M 5 = M2 + M 4 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. M1 + M 3 - M2 + M 6 A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-24 Formulas for Strassen’s Algorithm M1 = (A00 + A11) (B00 + B11) M2 = (A10 + A11) B00 M3 = A00 (B01 - B11) M4 = A11 (B10 - B00) M5 = (A00 + A01) B11 M6 = (A10 - A00) (B00 + B01) M7 = (A01 - A11) (B10 + B11) Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-25 Analysis of Strassen’s Algorithm If n is not a power of 2, matrices can be padded with zeros. What if we count both multiplications and additions? Number of multiplications: M(n) = 7M(n/2), M(1) = 1 Solution: M(n) = 7log 2n = nlog 27 ≈ n2.807 vs. n3 of brute-force alg. Algorithms with better asymptotic efficiency are known but they are even more complex and not used in practice. Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-26 Closest-Pair Problem by Divide-and-Conquer Step 0 Sort the points by x (list one) and then by y (list two). Step 1 Divide the points given into two subsets S1 and S2 by a vertical line x = c so that half the points lie to the left or on the line and half the points lie to the right or on the line. Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-27 Closest Pair by Divide-and-Conquer (cont.) Step 2 Find recursively the closest pairs for the left and right subsets. Step 3 Set d = min{d1, d2} We can limit our attention to the points in the symmetric vertical strip of width 2d as possible closest pair. Let C1 and C2 be the subsets of points in the left subset S1 and of the right subset S2, respectively, that lie in this vertical strip. The points in C1 and C2 are stored in increasing order of their y coordinates, taken from the second list. Step 4 For every point P(x,y) in C1, we inspect points in C2 that may be closer to P than d. There can be no more than 6 such points (because d ≤ d2)! Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-28 Closest Pair by Divide-and-Conquer: Worst Case The worst case scenario is depicted below: Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-29 Efficiency of the Closest-Pair Algorithm Running time of the algorithm (without sorting) is: T(n) = 2T(n/2) + M(n), where M(n) Θ(n) By the Master Theorem (with a = 2, b = 2, d = 1) T(n) Θ(n log n) So the total time is Θ(n log n). Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-30 Quickhull Algorithm Convex hull: smallest convex set that includes given points. An O(n^3) bruteforce time is given in Levitin, Ch 3. Assume points are sorted by x-coordinate values Identify extreme points P1 and P2 (leftmost and rightmost) Compute upper hull recursively: • find point Pmax that is farthest away from line P1P2 • compute the upper hull of the points to the left of line P1Pmax • compute the upper hull of the points to the left of line PmaxP2 Compute lower hull in a similar manner Pmax P2 P1 Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-31 Efficiency of Quickhull Algorithm Finding point farthest away from line P1P2 can be done in linear time Time efficiency: T(n) = T(x) + T(y) + T(z) + T(v) + O(n), where x + y + z +v <= n. • worst case: Θ(n2) T(n) = T(n-1) + O(n) • average case: Θ(n) (under reasonable assumptions about distribution of points given) If points are not initially sorted by x-coordinate value, this can be accomplished in O(n log n) time Several O(n log n) algorithms for convex hull are known Copyright © 2007 Pearson Addison-Wesley. All rights reserved. A. Levitin “Introduction to the Design & Analysis of Algorithms,” 2nd ed., Ch. 4 4-32